column number extract using find function..

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인수 송
인수 송 2022년 5월 6일
댓글: Jon 2022년 5월 6일
A = [ 0 500 1000 1500 2000 2500 3000 3500 x1 x2 x3 5500 6000 6500 7000 7500 8000 8500 9000 9500 10000 ] ;
i want to know the column number using find function about 500,1000,2000,3000, x2, 4000, 5000, 6000, 7000, 8000, 9000, 10000 ].
like above answer B
: B = [ 2 3 5 7 9 11 13 15 17 19 21 23 25 27 29 ] ; (example : it's not answer)
The x1, x2, x3 is variable value.

답변 (3개)

dpb
dpb 2022년 5월 6일
Use the optional second output form of find
[r,c]=find(....);
  댓글 수: 1
Jon
Jon 2022년 5월 6일
편집: Jon 2022년 5월 6일
I think this seems like it would only help you if you had a 2d array of logicals or 0's and 1's. The OP just has a 1d array. Unless maybe I am completely missing something. The issue is how to create the corresponding vector of logicals that find can be applied to. One way is using ismember as shown in my answer below.

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Jon
Jon 2022년 5월 6일
A = [ 0 500 1000 1500 2000 2500 3000 3500 x1 x2 x3 5500 6000 6500 7000 7500 8000 8500 9000 9500 10000 ] ;
matchVals = [500,1000,2000,3000, x2, 4000, 5000, 6000, 7000, 8000, 9000, 10000 ];
B = find(ismember(A,matchVals))
  댓글 수: 1
Jon
Jon 2022년 5월 6일
In your case the values seem to follow a very regular progression. I'm not sure if this is just the case for the example you give or if it always holds. If you have such a regular progression you could also write a formulae, e.g
B = matchVals/500 + 1

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Voss
Voss 2022년 5월 6일
[~,idx] = ismember([500 1000 2000],[0 500 1000 1500 2000 2500])
idx = 1×3
2 3 5
  댓글 수: 1
Jon
Jon 2022년 5월 6일
Good point, you don't need to use find as I did after calling ismember, just use the second output argument to get the indices

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